Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $n \neq 0$. $k = \dfrac{n - 1}{n^2 - 5n - 50} \div \dfrac{n - 1}{n^2 - 10n} $
Dividing by an expression is the same as multiplying by its inverse. $k = \dfrac{n - 1}{n^2 - 5n - 50} \times \dfrac{n^2 - 10n}{n - 1} $ First factor the quadratic. $k = \dfrac{n - 1}{(n - 10)(n + 5)} \times \dfrac{n^2 - 10n}{n - 1} $ Then factor out any other terms. $k = \dfrac{n - 1}{(n - 10)(n + 5)} \times \dfrac{n(n - 10)}{n - 1} $ Then multiply the two numerators and multiply the two denominators. $k = \dfrac{ (n - 1) \times n(n - 10) } { (n - 10)(n + 5) \times (n - 1) } $ $k = \dfrac{ n(n - 1)(n - 10)}{ (n - 10)(n + 5)(n - 1)} $ Notice that $(n - 1)$ and $(n - 10)$ appear in both the numerator and denominator so we can cancel them. $k = \dfrac{ n(n - 1)\cancel{(n - 10)}}{ \cancel{(n - 10)}(n + 5)(n - 1)} $ We are dividing by $n - 10$ , so $n - 10 \neq 0$ Therefore, $n \neq 10$ $k = \dfrac{ n\cancel{(n - 1)}\cancel{(n - 10)}}{ \cancel{(n - 10)}(n + 5)\cancel{(n - 1)}} $ We are dividing by $n - 1$ , so $n - 1 \neq 0$ Therefore, $n \neq 1$ $k = \dfrac{n}{n + 5} ; \space n \neq 10 ; \space n \neq 1 $